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Файл:Ординатура / Хирургия / Библиотека им академика М.И. Перельмана / Книга_155_библиотеки_им_акад_М_И_Перельмана
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Chapter 6
Normal Distribution
6.1 Normal Distribution
Suppose you buy a box of 16 hands disinfectious (200 ml each). When you put them
together at home, you do not see the liquid on the bottle in equal level. Of course,
just as you know it’s very unlikely that their net volume is exactly 200 cc, you can
never expect their volume to be too far from 200 cc. But you will not be surprised,
if each bottle contains a few cc more or less. To ensure their quantity, you measure
the volume of one. If it is only a few cc smaller in size, you will blame it on the
filling machine error, but you will probably go to the next bottle. If the volume of
disinfectious is more than 200 cc, you will not care. If the latter is larger and the
former is smaller, you may find these two bottles sufficient to measure. But if the
latter is as low as the former, your suspicious will increase and you will go to the
next bottles and write down their volume. Keep doing this until you reach a certain
and confidence result. If some bottles are smaller and some are larger enough to
compensate each other, it is justified that the average volume is the same 200 ml
“
200 ± 07 ml” written on them.
However, the presence of a 120–150 ml bottle between them is not acceptable to
you. The volume of disinfectant bottles is: (205–204–202–201–199–198–195–195–
194–193–188–190–211–209–208–206), and their chart is a bell shape and symmetric
(Fig.
6.1).
This pattern of bell-shaped distribution curves is called normal distribution.
Another name for this distribution, which is used for continuous variables, is the
Gaussian distribution. The larger the number of trials, the smoother, more
symmetrical, and bell-like this curve will be.
Many of the events we deal with in our daily lives, such as blood pressure, marital
age, maternal age, height, and weight, have an almost normal distribution. In the
medical and health sciences, too, many variables can be found that have relatively
normal or similar to normal distributions, or we have to assume that they are normal
until their abnormality is proven.
© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2024
S. H. Saneii and H. Doosti, Practical Biostatistics for Medical and Health Sciences,
https://doi.org/10.1007/978-981-97-3083-4_6
115
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116 6 Normal Distribution
Fig. 6.1 Result of repeated
weighing
of disinfectant
liquids
The density function of the normal distribution probability is as follows:
f
(
x
)
=
e
−
1
2
(
x−μ
)
σ
2
σ
√
2π
,
where π = 3.14 and e =. The normal distribution has two parameters μ and σ.By
specifying
these two parameters, the distribution pattern of the dataset is specified
and it is indicated by
X ∼ N
(
μ.
σ
)
.
6.1.1 Distribution Function and Density Function
Typically, every distribution is characterized by two functions: the density function,
denoted as f (x), and the distribution function, denoted as F(x). In the case of the
normal distribution, the density function, f (x), takes the form of a symmetrical
bell-shaped curve. The integral of this function over its entire domain equals one,
representing the total probability. Utilized to convey the distribution’s state and
structure, the density function is a fundamental aspect. On the other hand, the
distribution function, F(x), is primarily a cumulative probability function. It
enables us to easily determine the probability of observing data up to a specific
value of x, inclusive. Figure 3.5 illustrates both the distribution functions and the
density
function (Fig.
6.2).
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6.1 Normal Distribution 117
Fig. 6.2 Comparison between normal density and normal distribution functions
6.1.2 The Properties of the Normal Distribution
The normal distribution has important, obvious, and unique properties that are briefly
described below:
The normal distribution curve exhibits a characteristic bell shape with a single
peak. At the center of this distribution, the mean, median, and mode are all
identical, showcasing a central tendency. Because of its symmetric nature, the
normal distribution curve is unimodal, meaning it possesses only one mode. The
total area under the normal distribution curve is considered to be 100% or 1. Half of
this area (50%) lies to the right of the mean, while the remaining half (50%) is to
the left. The normal distribution is perpetually asymptotic, never intersecting the
X-axis but maintaining a parallel appearance to it. In essence, no matter how far the
curve extends in either direction, it never reaches the X-axis, yet gets progressively
closer to it. Observations occurring in the center of the distribution are more
frequent than those elsewhere, with frequency decreasing the further one moves
from the center. When defining the relative frequency of observations for a given
class with boundary points a and b, it is equivalent to the areas under the curve
between these two points. Exploiting the symmetry of this distribution and
employing the sample’s standard deviation estimation, we can deduce that within 1
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118 6 Normal Distribution
Fig. 6.3 Interpretation of
normal distribution in terms
of mean and standard
deviation
standard deviation of the mean, approximately 68% of the area under the curve is
encompassed; within 2 standard deviations, around 95% of the area is covered; and
within 3 standard deviations, about 99.7% of the area is contained.
Tips
The empirical rule for the normal distributions:
•
Within one standard deviation of the mean (μ − σ to μ + σ ) contains
approximately 68% of the population.
•
Within two standard deviations of the mean (μ − 2σ to μ + 2σ ) contains
approximately 95% of the population.
•
Within three standard deviations of the mean (μ − 3σ to μ + 3σ ) contains
approximately 99.7% of the population.
According to Chebyshev’s theorem, in any dataset, the proportion of data will fall
within K standard deviation of the mean (
x ± Kσ ) which is at least 1 −
1
K
2
, where K
is any positive number greater than 1. For example, at least
1 −
1
2
2
= 0.75 or 75%
of the data will be within 2 Sd of the mean and at least
1 −
1
3
2
= 0.89 or 89% of the
data will fall between 3 Sd of the mean (Fig.
6.3).
6.2 The Standard Normal Distribution
A special situation of a normal distribution whose mean and standard deviation are 0
and 1 is called the standard normal distribution N (0 and 1). Any normal distribution
can be transformed to a standard normal distribution to find the part of area under the
curve that lies between two points (interval) or to obtain the ratio of an area which
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6.2 The Standard Normal Distribution 119
is more or less than a point (a certain value). This transformation is done using the
following relation:
Z
i
=
X
i
− μ
σ
,
where μ is the mean, σ is the standard deviation, X
i
is a certain value of the
variable, and Z is the standardized variable. This procedure is called
standardization of frequency distribution because it can be used to make any
normal distribution equivalent to the normal standard distribution. The distance
from the mean under the curve of standard normal is the criterion for comparing
statistical tests and is called the Z criterion.
The standard normal distribution has another extra property (mean 0 and standard
deviation 1) in addition to all normal distribution properties. Based on this feature,
using the Z distribution table, the area under the curve of each normal variable can
be obtained.
To find the area under the curve, we first find the probability of Z regardless of its
sign from the table and then obtain it according to one of the following three basic
types:
Both points are on one side of the mean (both on the left side or both on the right
side). So find their probability from the standard normal table and subtract them.
Each of the two points on one side of the mean (one on the left side and the other
on the right side). After finding their probabilities from the standard normal table,
add them together.
When we are looking for the area under the curve to be more than one point or
less than one point, after finding the probability of that point (regardless of its sign)
from the standard normal table, subtract it from 0.5.
Tip
A normal distribution with the parameters of μ = 0 and σ = 1 is named the
standard normal distribution and denoted by
X ∼ N
(
0, 1
)
. The total area
under its density curve is equal to 1 or 100%.
Example 1 Suppose the IQ has a normal distribution with a mean of 100 and a
standard deviation of 15. What percentage of people has an IQ between 85 and 130?
Solution: First, we transform the above values to the standard normal variable using
the
Z
i
=
X
i
−μ
σ
formula.
p
(
85 ≤ x ≤ 130
)
= p
85 − 100
15
≤ z ≤
130 − 100
15
= p
(
−1 ≤ z ≤+2
)
.
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120 6 Normal Distribution
Fig. 6.4 Area between any two z values either side of the mean. Look up both corresponding values
to z and add together
Due to the symmetry property of the normal curve, the following relation is
established.
Due to the symmetry nature of the normal distribution curve and the fact that two
values are on either side of the mean value (one smaller and the other larger), the
above relation can be written as follows:
p
(
−1 ≤ z ≤+2
)
= p
(
−1 < z < 0
)
+ p(0 < z < 2).
Then using the standard normal area in the z table, we obtain the corresponding
probabilities.
p
(
−1 ≤ z ≤+2
)
= p
(
−1 < z < 0
)
+ p
(
0 < z < 2
)
= 0.3413 + 0.4772 = 0.8185.
That is, approximately 81.9% of people have an IQ between 85 and 130 (Fig. 6.4).
Example 2 In Example 1, the IQ had a normal distribution: IQ ∼ N
(
100, 15
)
. What
percentage of people has an IQ greater than 130?
Solution: First, we transform the value of 130 to the standard normal variable using
the Z
i
=
X
i
−μ
σ
formula.
p
(
x > 130
)
= p
z >
130 − 100
15
= p(z > 2).
The standard normal distribution table provides the area under the curve from the
mean (zero point) to a certain point. In order to obtain the area under the curve that
exceeds a certain point, i.e., to the end of the distribution, we must subtract the value
obtained from the table from 0.5. Because in normal distribution, the mean divides
the area under the curve into two equal parts.
p
(
x > 130
)
= p
(
z > 2
)
= 0.5 − p
(
z < 2
)
.
The probability of p(z < 2) from the standard normal distribution table is:
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6.2 The Standard Normal Distribution 121
Fig. 6.5 Area to the right of
any z value. Look up the
corresponding value to z and
subtract the area from 1
p
(
z < 2
)
= 0.4772.
And
p
(
x > 130
)
= p
(
z > 2
)
= 0.5 − p
(
z < 2
)
= 0.5 − 0.4772 = 0.0228.
That is, approximately 2.28% of people have an IQ more than 130 (Fig. 6.5).
Example 3 In Example 1, the IQ had a normal distribution: IQ ∼ N
(
100, 15
)
. What
percentage of people has an IQ less than 85?
Solution: First, we transform the value of 85 to the standard normal variable using
the
Z
i
=
X
i
−μ
σ
formula.
p
(
x < 85
)
= p
z <
85 − 100
15
= p(z < −1).
Since the normal distribution curve is symmetric, we can obtain p(z < +1) instead
of
p(z < −1).
p
(
−1 < z < 0
)
= p
(
0 < z < +1
)
= 0.3413.
In order to obtain the probability of being smaller than p(z < −1), we need to
reduce the above value from 0.5 (half left of the curve).
p
(
z < −1
)
= 0.5 − p
(
0 < z < +1
)
= 0.5 − 0.3413 = 0.1587.
That is, approximately 15.87% of people have an IQ less than 85 (Fig. 6.6).
p
(
x < 85
)
= p
z <
85 − 100
15
= p(z < −1).
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122 6 Normal Distribution
Fig. 6.6 Area to the left of
any z value. Look up the
value of z in the table and
subtract the area from 1
Example 4 In Example 1, the IQ had a normal distribution: IQ ∼ N
(
100, 15
)
. What
percentage of people has an IQ between 115 and 130?
Solution: Likewise, to previous examples, we transform the both values of 115 and
130 to the standard normal variable using the
Z
i
=
X
i
−μ
σ
formula.
p
(
115 ≤ x ≤ 130
)
= p
115 − 100
15
≤ z ≤
130 − 100
15
= p
(
+1 ≤ z ≤+2
)
.
The standard normal distribution table provides the area under the curve between
any specified points to the mean (0). Therefore, we must find the values corresponding
to the points 1 and 2.
p
(
0 < z < 2
)
= 0.4772,
p
(
0 < z < 1
)
= 0.3413.
Because both standardized values are positive, both are larger than the mean and
are on one side of the mean. Therefore, we have to subtract them to achieve the area
under the standard normal curve between values (1 and 2).
p
(
+1 ≤ z ≤+2
)
= p
(
0 < z < 2
)
−p
(
0 < z < 1
)
= 0.4772−0.3413 = 0.1359.
That is, approximately 13.59% of people have an IQ between 115 and 130
(Fig.
6.7).
Example 5 In Example 1, the IQ had a normal distribution: IQ ∼ N
(
100, 15
)
. What
percentage of people has an IQ between 80 and 90?
Solution: As previous example, we transform the both values of 80 and 90 to the
standard normal variable using the
Z
i
=
X
i
−μ
σ
formula.
p
(
80 ≤ x ≤ 90
)
= p
80−100
15
≤ z ≤
90−100
15
= p
(
−1.33 ≤ z ≤−0.67
)
.
From the standard normal distribution table and due to its symmetry, we obtain
the area under the curve for the values of 1.33 instead of −1.33 and 0.67 instead of
−0.67.
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6.2 The Standard Normal Distribution 123
Fig. 6.7 Area between any two z values at positive side of the mean. Look up both values of z in
the table and subtract corresponding areas
Fig. 6.8 Area between any two z values at negative side of the mean. Look up both values of z in
the table and subtract corresponding areas
p
(
0 < z < −1.33
)
= 0.3485,
p
(
0 < z < −0.67
)
= 0.2486.
Because both standardized values are negative, both are less than the mean and
are on one side of the mean. Therefore, we have to subtract them to achieve the area
under the standard normal curve between values −0.67 and −1.33.
p
(
−1.33 ≤ z ≤−0.67
)
= p
(
0 < z < −1.33
)
− p
(
0 < z < −0.67
)
= 0.3485 − 0.2486 = 0.0999.
That is, approximately 9.99% of people have an IQ between 80 and 90 (Fig. 6.8).
6.2.1 Skewness and Kurtosis of Normal Distributions
The other two properties of the normal distribution that specify the lack of symmetry
and peakedness of normal distributions are important.
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124 6 Normal Distribution
Skewness: Skewness is a measure of symmetry, or more precisely the lack of
symmetry. In normal distributions, in addition to being symmetric, the mean,
median, and mode are very close together. But sometimes when the mean, median,
and mode are not match, the distribution continued on one side (left or right) and is
longer on the other. This means that the location of mean and median changes,
while the location of mode appears to be unchanged in place. Therefore, the mean
and the median inclined to a larger and longer tail.
When a distribution continues from the right-hand side of the X-axis toward the
larger and positive values, the tail of that distribution is more elongated from the
right, and the median and mean are larger than the mode of the distribution, and
that distribution is called “skew to the right” or “positive skewness (such as the
community income). But if the distribution continues to the left of the X-axis, the tail
of that distribution extends to the l eft, that is, to the lesser and negative values, and
that distribution is called “skew to the left” or “negative skewness” (age at death).
In addition to visually detecting of skewness, three measures for calculating the
skewness of a distribution are given below, which is known as the Pearson coefficient,
the Galton coefficient, and the third moment.
Sk =
3
(
μ − Med
)
σ
or SK =
Q
3
− 2Q
2
+ Q
1
Q
3
− Q
1
.
These coefficients are unitless. Because in exactly symmetric and normal
distributions, the mean and the median coincide, these measures must be equal to
zero. Mathematically, these coefficients do not have upper and lower limits. The
closer these values to zero, the closer the distribution is to normal. They rarely even
come close to +1 and −1. In any case, if
Sk < 0, the skewness is negative, and if
Sk > 0, the skewness is positive.
The G-coefficient is another measure for detecting symmetry (skewness) of the
distribution:
G =
1
n
×
(
x
i
− x
)
3
σ
3
.
In normal distribution, this G = 0. If this coefficient:
•
Less than 0.1 (
|
G
|
≤ 0.1
), the distribution is assumed to be almost normal.
•
Between 0.1 and 0.5 (0.1 ≤
|
G
|
≤ 0. 5) Low distribution skewness.
•
Greater than 0.5 (
|
G
|
≥ 0.5
), the distribution is considered to be a highly
distributed skew.
It should be noted that statistical software such as SPSS and SAS calculates
skewness for the sample, whose formula is slightly different (Fig.
6.9).
G =
n
(
n − 1
)(
n − 2
)
×
(
x
i
− x
)
3
σ
3
.
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